Question

EDO - Resolva o problema de valor inicial:

$\frac{dy}{dx}=\frac{x}{(x^2+9)^{\frac{1}{2}}}$, $y(4)=2$.

Se puder passar o passo a passo, fico grato.

Answer 1

Boa noite Mmedilson!

Solução!

Vamos aplicar a integral para resolver o P. V. I.

$\displaystyle\int \frac{x}{( x^{2} +9)^{ \frac{1}{2} } } \\\\\\\ Fazendo~~por ~~substituic\~ao!\\\\\\\\ u= x^{2} +9\\\\\ du=2xdx\\\\\\ \boxed{\frac{du}{2}=xdx}$


$\displaystyle\int \frac{1}{2 \sqrt{u} } } \\\\\\\ \dfrac{1}{2} \displaystyle\int \frac{1}{\sqrt{u} } } \\\\\\\ \dfrac{1}{2} \displaystyle\int \frac{1}{\sqrt{x} } } \\\\\\\ \dfrac{1}{2} \displaystyle\int \frac{1}{(x)^{ \frac{1}{2} } } } \\\\\\\ \dfrac{1}{2} \displaystyle\int (x)^{ -\frac{1}{2} } } } \\\\\\\ \dfrac{1}{2} \displaystyle\int \frac{ (x)^{ -\frac{1}{2}+1 } }{- \frac{1}{2} +1} } } \\\\\\\ \dfrac{1}{2} \displaystyle\int \frac{ (x)^{\frac{1}{2} } }{ \frac{1}{2}}$


$\displaystyl (x)^{\frac{1}{2}}\\\\\\ x=u\\\\\\ \displaystyle\int \frac{x}{( x^{2} +9)^{ \frac{1}{2} } }=\sqrt{ x^{2} +9}+c$

Agora vamos igualar a função integrada a esses valores,para determinarmos o valor da constante.


$y(4)=2 y=\sqrt{ x^{2} +9}+c\\\\\\\ 2=\sqrt{ (4)^{2} +9}+c\\\\\\ 2=\sqrt{ 25}+c\\\\\\ 2=5+c\\\\\\ 2-5=c\\\\\ -3=c\\\\\\ \boxed{c=-3}$

$y=\sqrt{ x^{2} +9}+c\\\\\\ y= \sqrt{ x^{2} +9}-3\\\\\\\\ \boxed{Resposta:~~y=\sqrt{ x^{2} +9}-3}$

Boa noite!
Bons estudos!